Laboratory Tensile Test

Open AccessLaboratory

Tensile TestMaterials and Electrical Science

Thomas Reason 10218419

Amit Ramji 10241445

Tinu Raju 09242606

Edward Raprager 10230101

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Contents Page

Content Page number

Background Research 3

Theory and Methodology 5

Test Procedure 6

Results and Analysis 7

Conclusion 11

10 Questions 12

References 15

Appendix 1 – Steel Data

Appendix 2 – Aluminium Data

Appendix 3 – Polypropylene Data

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Structural Background Research

Tensile tests are carried out for a number of reasons. The mainreason is for engineering applications. When a designer orstructural engineer designs a new product, they have to be able toselect which materials will be best suited for the job. It could beunsafe to use a material that is not strong enough for theapplication. It is for this reason that tensile tests are carriedout.

Materials are usually chosen so that the maximum stress that thematerial will encounter is roughly half of the stress at the yieldpoint. This is to ensure that there is a margin of safety so thematerial won’t fail.

Steel and Aluminium

Atoms in steel and aluminium are bonded in metallic structures. Thiscan be described as a sea of delocalised electrons surrounding positively charged nuclei. This is why metals are good conductors ofheat and electricity. (Shackelford J E 2005 1)

The steel has carbon interstitial impurities in between iron atoms. This has the effect of increasing the yield and ultimate strengths. It acts as an alloying element. When a metal undergoes a stress, themetal atoms want to slip over the top of each other. This causes plastic deformation. The atoms will slip in slip planes, as less force is needed to move one atom at a time than it is to move many atoms at one. The alloying element of carbon reduces the ability of the atoms to move and therefore increases the strength.

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The aluminium alloy that will be tested is made from several key elements. Aluminium makes up 95-98% of the alloy with magnesium being the next biggest element by percentage at 1.6-2.2% (MatWeb 2).Carbon and magnesium are very similar in terms of their atomic radius, and therefore the magnesium will be a substitution impurity atom.

Polypropylene

Polypropylene will show a low values for the strengths due to the way that the atoms are bonded. The material is made up of chains of hydrocarbons which are covalently bonded. These covalent bonds are relatively strong, but because the hydrocarbon chains are not uniform in direction they have the ability to stretch into straight chains. This means that they can undergo relatively large extensionswithout plastically deforming, but due to the weak interactions between the chains it is ultimately not as strong as either of the two metals that will be tested.

Aluminum Alloy

Aluminum alloy is malleable and can be bent into shapes very easily although it lacks strength under heavy loads. We expect to see a steep line initially while it is still elastic but then a sharp dropoff after then quickly breaking. We don't expect the test material to show any signs of stretching as once it has become plastic it will most likely break. The atoms are arranged in face centered cubic structure.

(MatWeb 3)

Mild Steel

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The carbon atoms interstitial impurities within the steel structure.http://clearscience.tumblr.com/post/628686089/steel-is-not-an-element-

Average Value

Young’s Modulus 70 GPa

Yield Stress 242 MPa

Tensile Strength 303 MPa

Steel has a much higher density than aluminum and is much more toughbut less malleable. We’d expect to see a slightly shallower line where it is still elastic but then will slowly start to deform. Its high tensile strength will mean that it will undergo elongation before it breaks. The atoms are arranged in a Body Centered Cubic structure.

(MatWeb 4)Polypropylene

We expect to see a very shallow line initially. It will very quicklydeform and stretch. There will be a large region of plastic deformation.

(MatWeb 5)

Theory of Tensile Test

The objective of Tensile Test is to obtain the Ultimate Tensile Strength, Young’s Modulus and Yield Strength. The Tensile test is probably the best way to determine how materials will react to the forces being applied in tension. They are easy to carry out and are not expensive once the test machine has been obtained. The specimen is pulled in opposite directions until it breaks, and the relationship between stress, which is derived from the load applied on the material, and strain, derived from the elongation of the sample, is given by a Stress-Strain curve.

Methodology

The length, width and thickness of the sample materials are measuredbefore starting the test. At the start the crosshead of the Tensile

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Average Value

Young’s Modulus 203 GPa

Yield Stress 689 MPa


Average Value

Young’s Modulus 1.8 GPa

Yield Stress 32.8 MPa


Test machine should be at the initial position. The specimens are gripped within the jaws of the machine and the load is gradually increased on the sample materials. The readings are recorded by the computer while the specimens are being extended. The specimen breaksat a certain point and that point is called the breaking Point. The sample materials are removed once they are broken, and the new lengths are measured. The tensile stress and tensile strain are calculated and the stress-strain graph is plotted which helps in finding the young’s modulus and the yield strength.

The test procedure

To carry out the test, the 10KN Hounsfield tensile test machine was used. The width and thickness of the test material was measured as

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this is used to calculate the cross sectional area of the test piece, in millimetres.

CrosssectionalArea=WidthxThickness

The length of the test material was also measured as this is needed later to calculate the strain. The test machine is linked up to a computer that logs the raw load and the raw displacement of the cross head. By setting up the machine in a suitable way the test canbe carried out.

For steel and aluminium, the test machine to give a force range of 100% with a cross head speed of 10mm per millimetre, with a 100mm extension. Polypropylene used a force range of only 10% and had a cross head speed of 100mm per minute. The extension was set to 1000mm.

By doing some post processing of the data, the stress and strain canbe calculated. This will enable the Young’s modulus to be calculatedand be able plot a graph so the yield point, maximum tensile strength and strength at failure can be determined.

Stess=ForceArea Where force is measured in Newtons and area is

measured in mm2.

Strain=Extension

OriginalLenght=l−l0

l0

Where L is the final length and L0 is the original length measured in mm.

The young’s modulus of the material can be calculated by using the values of stress and strain.

Young'sModulus(MPa)=StressStrain

The maximum tensile strength can be read off a stress strain graph as the highest stress value. Alternatively, maximum tensile strength

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can be calculated by dividing the maximum applied load by the cross sectional area.

Maximumtensilestrength= MaximumloadCrosssectionalarea

=StressxCSACSA

It can be seen how the maximum tensile strength is equal to the maximum value for stress.

Results for tensile test on Steel

The graph below shows the stress strain curve for mild steel (CD01/CR4).

The graph can be used to calculate the Young’s Modulus, Yield Strength, maximum tensile strength, and failure strength.

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0 0.02 0.04 0.06 0.08 0.1 0.12 0.140

50

100

150

200

250

300

350

Stress strain curve for Steel

Strain

Stress (N/mm)

To calculate the Young’s Modulus the linear part of the graph can beused. Young’s Modulus is calculated by dividing stress by strain. The graph below focuses on the linear section.

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02050100150200250300

Stress strain curve for Steel

Strain

Stre

ss (N/

mm)

From the graph we can see that the linear region is between 0 and 150 in the Y axis. As the Young’s Modulus is stress divided strain we can calculate that the Young’s modulus for this specimen of steelis

Young'sModulus=Stressstrain

=1500.004

=37500MPa=37.5GPa

The maximum tensile strength is calculated by dividing the maximum load by the cross sectional area.

Maximumtensilestrength= 59671.59x12.51

=300MPa

The yield strength is the maximum force that can be applied to the material without plastically deforming. On a graph, this can be seenas the point where the linear section becomes non-linear. At the yield, point the load applied at that point can be calculated by multiplying the stress by the cross sectional area. At a stress of 150 MPa the load on the specimen was 2985 Newtons. The yield strength can be calculated by dividing this force by the cross sectional area.

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YieldStrength= 29851.59x12.51

=150MPa

The strength at failure is calculated in a similar way, by dividing the load at the point of failure by the cross sectional area.

Strengthatfailure=183x(1.59x12.51)1.59x12.51

=183MPA

Results for tensile test on Aluminium

The graph below shows the stress strain curve for Aluminium 5251

H22.

Like steel, the graph can be used to calculate the Young’s Modulus, Yield Strength, maximum tensile strength and failure strength.

The graph below will help to calculate the Young’s Modulus more easily.

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0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.040

50

100

150

200

250

Stress Strain Curve for Aluminium

Strain

Stre

ss

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

50

100

150

200

250

Stress Strain Curve for Aluminium

Strain

Stress

The graph shows a linear region between the stress values of 0 and 180 MPa.


=180

0.00472=38135MPa=38.1GPa


Maximumtensilestrength=4180

1.44x12.55=231MPa

The yield strength is the maximum force that can be applied to the material without plastically deforming. On a graph, this can be seenas the point where the linear section becomes non-linear. At the yield point, the load applied at that point can be calculated by multiplying the stress by the cross sectional area. At a stress of 180 MPa the load on the specimen was 3275 Newtons. The yield strength can be calculated by dividing this force by the cross sectional area.

YieldStrength=3275

1.44x12.55=181MPa


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Strengthatfailure=185x(1.44x12.55)1.44x12.55

=185MPA

Results for tensile test on Polypropylene

The graph below shows the stress strain curve for Natural COPOL Polypropylene.7

The graph below will focus on the linear section of the graph. This will help to determine the yield point and Young’s Modulus more easily.

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0 0.02 0.04 0.06 0.08 0.1 0.12 0.140

20

40

60

80

100

Stress Strain Graph for Polyproplyene

Strain

Stre

ss Mpa

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.0040

10

20

30

40

50

60

70

80

90

Stress Strain Graph for Polyproplyene

Strain

Stress Mpa

The graph shows a linear region between the stress values of 0 and 64MPa.


= 640.0005

=128000MPa=128GPa


Maximumtensilestrength= 159612.59x1.49

=85MPa

The yield strength is the maximum force that can be applied to the material without plastically deforming. On a graph, this can be seenas the point where the linear section becomes non-linear. At the yield point, the load applied at that point can be calculated by multiplying the stress by the cross sectional area. At a stress of 64MPa the load on the specimen was 1200.6 Newtons. The yield strength can be calculated by dividing this force by the cross sectional area.

YieldStrength= 1200.61.49x12.59

=64MPa


Strengthatfailure=74.4x(1.49x12.59)

1.49x12.59=74.4MPA

Conclusion

The result of the above experiment is that we were able to obtainthe Young’s Modulus, Yield Strength, Ultimate Tensile Strength,Strength at Failure and the stress – strain curves of the materialswe have tested.

The average tensile strength of Mild Steel is approximately 785Mpa.We were expecting to see a higher tensile strength in the testspecimen with a strength ranging from 250MPa to 2400MPa depending onthe specific carbon content. It can be seen that a value of 300MPawas recorded, which falls within the expected range.

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The tensile strength of pure Aluminium is normally less than that ofsteel and it is approximately 90Mpa. The specimen we have tested isan aluminium alloy and the tensile strength of pure aluminium isincreased by the addition of alloying elements. Hence, we wereexpecting to record a higher tensile strength in the given specimenand as expected we recorded a higher strength of 231Mpa.

Polypropylene has a lower tensile strength compared to other twomaterials we are testing. It normally ranges from 29MPa to 37MPa.However, we have calculated a value of 64MPa which is slightlyhigher than the normal case. We can assume that this may be becauseof the difference in the material we have considered for the test.

We have noticed from the above tests that the tensile strength ofMild Steel is 300MPa and it has the highest tensile strength amongthe other materials that we tested and Polypropylene has the lowesttensile strength.

1.Calculate the cross-sectional area of the samples:Average Thickness (mm)

Average Width (mm)

Average cross sectional area (mm2)

Steel 1.59 12.51 19.89Aluminum Alloy 1.44 12.55 18.07Polypropylene (PP) 1.49 12.59 18.76

2. Record the maximum load and stress for each materialsMax Load (N) Stress (N/mm2 = N.mm-

2)Steel 5950 299.1Aluminum Alloy 4180 231.3Polypropylene (PP) 1596.5 85.11

3. Calculate the maximum tensile stress for steel and Al-alloy and PP:Stress = Force/Areaσ= Fmax/AσSteel=5950N / 19.89mm = 299.1N/mm2

σAluminum=4180N / 18.07mm = 231.3 N/mm2

σpolypropylene=1596.5N / 18.76mm = 85.11 N/mm2

4. Record amount by which the sample has extended in mm:Original Length (mm) [L0]

Extended Length (mm)

e = ∆L = L1 - L0 (mm)

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[L1]Steel 205 230 25Aluminum Alloy 212 7Polypropylene (PP) 300 95

5. Calculate the strain at fracture for 3 materials:Steel: Load = 3640N, Strain = 0.116Aluminum: Load = 3340N, Strain = 0.031Polypropylene (PP): Load = 1395N, Strain = 0.410

6. From your plot estimate the yield stress.Steel: 170.6 N/mm2

Aluminum Alloy: 175.9 N/mm2

Polypropylene (PP): 83.8 N/mm2

7. From your plot calculate the Young’s modulus.

E=σϵ =

σ2−σ1ϵ2−ϵ1 [In the elastic region]

ESteel = 45,100 N/mm2

EAluminum = 23,900 N/mm2

EPP = 530 n/mm8. Indicate on your plot the elastic and plastic regions. Explain the difference between these two terms.

At the origin there is no initial stress or strain the sample endures. Up to the proportional limit, the material follows Hooke’s Law; this is the linear region on the graph known as the elastic region. The material can then regain its original shape after removal of the load.

The proportion of the curve between the elastic limit or yieldpoint is known as the plastic region. The section beyond the elasticregion (linear on the graph) is known as the plastic regions where the cross sectional area of the material starts to decrease and the stress decreases where the test piece changes in length. This changein length and reduction of the cross sectional area continues with little increase in stress in the metallic materials until fracture or ultimate/maximum tensile stress.

The graphs below indicate the regions of elastic and

Area of elastic deformation Area of plastic deformation

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Only the first part of the graph has been shown above, as otherwise the linear region would not be distinguishable.

9. If the area of the mild steel is doubled, calculate the load thatis required to apply to it to reach the maximum tensile stress.

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Area of mild steel= 19.89mm2 , σ= Fmax/A , σSteel=5950N/19.89mm = 299.1N/mm2

Proposed Area = 2x19.89 = 39.78mm2

Then: Fmax = A x σ = 39.78mm2 x 299.1N/mm2 = 11898 N ≈ 11.9KN

10. Give 3-5 part applications of each material in aerospace (i.e. A380):

Steel: Structural uses such as undercarriage, spars, struts, stringers, fixtures and fittings such as hinges, screws and nuts andbolts and many others.

Aluminum Alloy: Aircraft skin, wings, passenger seats and floor panels, control surfaces such as ailerons and rudders and manyother areas.

Polypropylene: Flexible fixtures such as washers, components of seats, housing compartments, wheels of food carts and more.

References

Reference Number

Page number

Reference Source

1 3 Shackelford J E, 2005, Material Science for Engineers, 6th Edition, Pearson

2 3 Matweb.com http://www.matweb.com/search/DataSheet.aspx?MatGUID=dbd6d7483df7481bbfa96475bc9fd1d8Aluminium Association

3 4 www.Matweb.com Material Property Website - Online database of engineering materials – 5000 series non ferrous metals.

4 4 www.Matweb.com Material Property Website

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http://www.matweb.com/search/DataSheet.aspx?MatGUID=dbd6d7483df7481bbfa96475bc9fd1d8

http://www.matweb.com/search/DataSheet.aspx?MatGUID=dbd6d7483df7481bbfa96475bc9fd1d8

- Online database of engineering materials5 4 www.Matweb.com Material Property Website

- Online database of engineering materials

Other information from Lecture notes and slides by Dr Ren at the University of Hertfordshire.

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Laboratory Tensile Test

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